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  <meta name="description" content="首先要知道指数的概念，如果对于一个数的k指数次幂满足模p同余1，那么这个最小正整数k就是指数，指数很好求，快速幂直接求得。 那么原根即k&#x3D;phi(p)即可，所以求原根有两种方法，第一种就是暴力枚举，具体定理可参考： https:&#x2F;&#x2F;www.cnblogs.com&#x2F;ldysy2012&#x2F;p&#x2F;12208905.html 证明什么的很详细，无非就是验证一下p-1质因数分解之后，计算一下公式中是否与1同余，">
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<meta property="og:description" content="首先要知道指数的概念，如果对于一个数的k指数次幂满足模p同余1，那么这个最小正整数k就是指数，指数很好求，快速幂直接求得。 那么原根即k&#x3D;phi(p)即可，所以求原根有两种方法，第一种就是暴力枚举，具体定理可参考： https:&#x2F;&#x2F;www.cnblogs.com&#x2F;ldysy2012&#x2F;p&#x2F;12208905.html 证明什么的很详细，无非就是验证一下p-1质因数分解之后，计算一下公式中是否与1同余，">
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          数论-原根
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        <p>首先要知道指数的概念，如果对于一个数的k指数次幂满足模p同余1，那么这个最小正整数k就是指数，指数很好求，快速幂直接求得。</p>
<p>那么原根即k=phi(p)即可，所以求原根有两种方法，第一种就是暴力枚举，具体定理可参考：</p>
<p><a target="_blank" rel="noopener" href="https://www.cnblogs.com/ldysy2012/p/12208905.html">https://www.cnblogs.com/ldysy2012/p/12208905.html</a></p>
<p>证明什么的很详细，无非就是验证一下p-1质因数分解之后，计算一下公式中是否与1同余，只要有一个与1同余就不满足。第二种方法即先求一个原根，根据定理，(d,p-1)=1满足，即d遍历p-1的所有简化剩余系，如果求得的一个原根为g，那么g^d mod p即为原根集合，原根个数为phi(phi(p-1))。</p>
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class="meta-string">&lt;vector&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string">&lt;ctime&gt;</span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">using</span> <span class="keyword">namespace</span> <span class="built_in">std</span>;</span><br><span class="line"><span class="keyword">using</span> ll = <span class="keyword">long</span> <span class="keyword">long</span>;</span><br><span class="line"></span><br><span class="line"><span class="keyword">clock_t</span> start1,end1,start2,end2;</span><br><span class="line"><span class="keyword">double</span> total1,total2;</span><br><span class="line"></span><br><span class="line"><span class="comment">//最小原根求解qx</span></span><br><span class="line"><span class="built_in">vector</span>&lt;ll&gt; qx;</span><br><span class="line"><span class="comment">//素因子</span></span><br><span class="line"><span class="built_in">vector</span>&lt;ll&gt; factor;</span><br><span class="line"><span class="comment">//简化剩余系</span></span><br><span class="line"><span class="built_in">vector</span>&lt;ll&gt; rest;</span><br><span class="line"></span><br><span class="line"><span class="comment">//a^k mod p</span></span><br><span class="line"><span class="function">ll <span class="title">fastpow</span><span class="params">(ll a,ll k,ll p)</span></span>&#123;</span><br><span class="line">    ll res = <span class="number">1</span>;</span><br><span class="line">    <span class="keyword">while</span>(k)&#123;</span><br><span class="line">        <span class="keyword">if</span>(k&amp;<span class="number">1</span>) res = (ll) res*a%p;</span><br><span class="line">        a = (ll)a*a%p;</span><br><span class="line">        (ll)k &gt;&gt;= <span class="number">1</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> res;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line"><span class="function">ll <span class="title">gcd</span><span class="params">(ll a,ll b)</span></span>&#123;</span><br><span class="line">    ll res;</span><br><span class="line">    res = a%b;</span><br><span class="line">    <span class="keyword">while</span>(res)&#123;</span><br><span class="line">        a=b,b=res;</span><br><span class="line">        res = a%b;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> b;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span></span>&#123;</span><br><span class="line">    ll p;</span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt; <span class="string">&quot;输入一个奇素数p&quot;</span> &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line">    <span class="built_in">cin</span> &gt;&gt; p;</span><br><span class="line">    ll phi = p<span class="number">-1</span>;</span><br><span class="line"></span><br><span class="line">    <span class="built_in">cout</span>&lt;&lt;<span class="string">&quot;以下为先求一原根再求全部原根过程&quot;</span>&lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line">    start1=clock();</span><br><span class="line">    start2=clock();</span><br><span class="line">    <span class="keyword">for</span>(ll i = <span class="number">2</span>; i &lt;= phi/i ; ++i)&#123;</span><br><span class="line">        <span class="keyword">if</span>(phi % i == <span class="number">0</span>) &#123;</span><br><span class="line">            factor.push_back(i);</span><br><span class="line">            <span class="keyword">while</span>(phi%i==<span class="number">0</span>) phi/=i;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">if</span>(phi&gt;<span class="number">1</span>) factor.push_back(phi);</span><br><span class="line"></span><br><span class="line">    end1=clock();</span><br><span class="line">    end2=clock();</span><br><span class="line">    total1=total1+(<span class="keyword">double</span>)(end1-start1)/CLOCKS_PER_SEC;</span><br><span class="line">    total2=total2+(<span class="keyword">double</span>)(end2-start2)/CLOCKS_PER_SEC;</span><br><span class="line">    phi = p - <span class="number">1</span>;</span><br><span class="line"></span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt; <span class="string">&quot;p-1分解质因数为:&quot;</span> &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">auto</span> x:factor)&#123;</span><br><span class="line">        <span class="built_in">cout</span> &lt;&lt; x &lt;&lt; <span class="string">&quot; &quot;</span> ;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line"></span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">auto</span> x:factor)</span><br><span class="line">        qx.push_back(phi/x);</span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt;<span class="string">&quot;应验证指数为:&quot;</span> &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">auto</span> x:qx)</span><br><span class="line">        <span class="built_in">cout</span> &lt;&lt; x &lt;&lt; <span class="string">&quot; &quot;</span> ;</span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line"></span><br><span class="line">    ll g=<span class="number">0</span>;</span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt; <span class="string">&quot;其中一个原根为:&quot;</span> &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line">    start1=clock();</span><br><span class="line">    <span class="keyword">for</span>(ll x=<span class="number">2</span>;x&lt;p;x++)&#123;</span><br><span class="line">        ll cnt=<span class="number">0</span>;</span><br><span class="line">        <span class="keyword">while</span> (cnt&lt;qx.size())&#123; </span><br><span class="line">            <span class="keyword">if</span>(fastpow(x,qx[cnt],p)==<span class="number">1</span>) <span class="keyword">break</span>;</span><br><span class="line">            cnt++;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">if</span>(cnt == qx.size())&#123;</span><br><span class="line">            g = x;</span><br><span class="line">            <span class="keyword">break</span>;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    end1=clock();</span><br><span class="line">    total1=total1+(<span class="keyword">double</span>)(end1-start1)/CLOCKS_PER_SEC;</span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt; g &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line"></span><br><span class="line">    start1=clock();</span><br><span class="line">    <span class="keyword">for</span>(ll i=<span class="number">1</span>;i &lt; phi; ++i)&#123;</span><br><span class="line">        <span class="keyword">if</span>(gcd(phi,i)==<span class="number">1</span>) </span><br><span class="line">            rest.push_back(i);</span><br><span class="line">    &#125;</span><br><span class="line">    end1=clock();</span><br><span class="line">    total1=total1+(<span class="keyword">double</span>)(end1-start1)/CLOCKS_PER_SEC;</span><br><span class="line"></span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt; <span class="string">&quot;p-1的简化剩余系为:&quot;</span> &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">auto</span> x:rest)</span><br><span class="line">        <span class="built_in">cout</span> &lt;&lt; x &lt;&lt; <span class="string">&quot; &quot;</span>;</span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt; <span class="string">&quot;原根的个数为:&quot;</span> &lt;&lt; rest.size() &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line">    </span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt; <span class="string">&quot;模p的所有原根为:&quot;</span> &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line">    </span><br><span class="line">    start1=clock();</span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">auto</span> &amp;x:rest)&#123;</span><br><span class="line">        <span class="built_in">cout</span> &lt;&lt; fastpow(g,x,p) &lt;&lt; <span class="string">&quot; &quot;</span> ;</span><br><span class="line">    &#125;</span><br><span class="line">    end1=clock();</span><br><span class="line">    total1=total1+(<span class="keyword">double</span>)(end1-start1)/CLOCKS_PER_SEC;</span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt;<span class="built_in">endl</span>;</span><br><span class="line"></span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt; <span class="string">&quot;***************************************&quot;</span> &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt;<span class="string">&quot;先求一个原根再求全部原根的时间为:&quot;</span> &lt;&lt; total1 &lt;&lt;<span class="string">&quot;s&quot;</span> &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line"></span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt; <span class="string">&quot;***************************************&quot;</span> &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt; <span class="string">&quot;暴力枚举全部原根的过程为:&quot;</span> &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line"></span><br><span class="line">    start2=clock();</span><br><span class="line">    <span class="keyword">for</span>(ll x=<span class="number">2</span>;x&lt;p;x++)&#123;</span><br><span class="line">        ll cnt=<span class="number">0</span>;</span><br><span class="line">        <span class="keyword">while</span> (cnt&lt;qx.size())&#123; </span><br><span class="line">            <span class="keyword">if</span>(fastpow(x,qx[cnt],p)==<span class="number">1</span>) <span class="keyword">break</span>;</span><br><span class="line">            cnt++;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">if</span>(cnt == qx.size())&#123;</span><br><span class="line">            g = x;</span><br><span class="line">            <span class="built_in">cout</span> &lt;&lt; g &lt;&lt; <span class="string">&quot; &quot;</span> ;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line">    end2=clock();</span><br><span class="line">    total2=total2+(<span class="keyword">double</span>)(end2-start2)/CLOCKS_PER_SEC;</span><br><span class="line"></span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt; <span class="string">&quot;***************************************&quot;</span> &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line">    <span class="built_in">cout</span> &lt;&lt;<span class="string">&quot;暴力枚举所有原根的时间为:&quot;</span> &lt;&lt; total2 &lt;&lt;<span class="string">&quot;s&quot;</span> &lt;&lt; <span class="built_in">endl</span>;</span><br><span class="line">    <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>


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